If $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is :
Answer & explanation
Correct answer: option 1
Put x = $\tan \theta$
∴ $y=\sin ^{-1}(\cos 2 \theta)=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2 \theta\right)\right)$
$=\frac{\pi}{2}-2 \theta=\frac{\pi}{2}-2 \tan ^{-1} x$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^2}$
Hence (1) is correct answer.